给定一个N叉树，任务是以图形方式打印 N 叉树。
树的图形表示：树的表示，其中根打印在一行中，子节点打印在后续行中，并带有一定量的缩进。
例子：
Input:
0
/ | \
/ | \
1 2 3
/ \ / | \
4 5 6 7 8
|
9
Output:
0
+--- 1
| +--- 4
| +--- 5
+--- 2
+--- 3
+--- 6
+--- 7
| +--- 9
+--- 8
方法：想法是使用 DFS Traversal 遍历 N-ary Tree 来遍历节点并探索其子节点，直到访问完所有节点，然后类似地遍历兄弟节点。
上述方法的分步算法如下所述 –
- 初始化一个变量来存储节点的当前深度，对于根节点，深度为 0。
- 声明一个布尔数组来存储当前的探索深度，并将它们全部标记为 False。
- 如果当前节点是根节点，即节点深度为0，则只需打印该节点的数据。
- 否则，循环从 1 到节点的当前深度并打印，’|’每个探索深度和非探索深度的三个空格只打印三个空格。
- 打印节点的当前值并将输出指针移动到下一行。
- 如果当前节点是该深度的最后一个节点，则将该深度标记为非探索。
- 同样，使用递归调用探索所有子节点。
下面是上述方法的实现：
C++
// C++ implementation to print
// N-ary Tree graphically
#include
#include
#include
using namespace std;
// Structure of the node
struct tnode {
int n;
list root;
tnode(int data)
: n(data)
{
}
};
// Function to print the
// N-ary tree graphically
void printNTree(tnode* x,
vector flag,
int depth = 0, bool isLast = false)
{
// Condition when node is None
if (x == NULL)
return;
// Loop to print the depths of the
// current node
for (int i = 1; i < depth; ++i) {
// Condition when the depth
// is exploring
if (flag[i] == true) {
cout << "| "
<< " "
<< " "
<< " ";
}
// Otherwise print
// the blank spaces
else {
cout << " "
<< " "
<< " "
<< " ";
}
}
// Condition when the current
// node is the root node
if (depth == 0)
cout << x->n << '\n';
// Condition when the node is
// the last node of
// the exploring depth
else if (isLast) {
cout << "+--- " << x->n << '\n';
// No more childrens turn it
// to the non-exploring depth
flag[depth] = false;
}
else {
cout << "+--- " << x->n << '\n';
}
int it = 0;
for (auto i = x->root.begin();
i != x->root.end(); ++i, ++it)
// Recursive call for the
// children nodes
printNTree(*i, flag, depth + 1,
it == (x->root.size()) - 1);
flag[depth] = true;
}
// Function to form the Tree and
// print it graphically
void formAndPrintTree(){
int nv = 10;
tnode r(0), n1(1), n2(2),
n3(3), n4(4), n5(5),
n6(6), n7(7), n8(8), n9(9);
// Array to keep track
// of exploring depths
vector flag(nv, true);
// Tree Formation
r.root.push_back(&n1);
n1.root.push_back(&n4);
n1.root.push_back(&n5);
r.root.push_back(&n2);
r.root.push_back(&n3);
n3.root.push_back(&n6);
n3.root.push_back(&n7);
n7.root.push_back(&n9);
n3.root.push_back(&n8);
printNTree(&r, flag);
}
// Driver Code
int main(int argc, char const* argv[])
{
// Function Call
formAndPrintTree();
return 0;
}
Java
// Java implementation to print
// N-ary Tree graphically
import java.util.*;
class GFG{
// Structure of the node
static class tnode {
int n;
Vector root = new Vector<>();
tnode(int data)
{
this.n = data;
}
};
// Function to print the
// N-ary tree graphically
static void printNTree(tnode x,
boolean[] flag,
int depth, boolean isLast )
{
// Condition when node is None
if (x == null)
return;
// Loop to print the depths of the
// current node
for (int i = 1; i < depth; ++i) {
// Condition when the depth
// is exploring
if (flag[i] == true) {
System.out.print("| "
+ " "
+ " "
+ " ");
}
// Otherwise print
// the blank spaces
else {
System.out.print(" "
+ " "
+ " "
+ " ");
}
}
// Condition when the current
// node is the root node
if (depth == 0)
System.out.println(x.n);
// Condition when the node is
// the last node of
// the exploring depth
else if (isLast) {
System.out.print("+--- " + x.n + '\n');
// No more childrens turn it
// to the non-exploring depth
flag[depth] = false;
}
else {
System.out.print("+--- " + x.n + '\n');
}
int it = 0;
for (tnode i : x.root) {
++it;
// Recursive call for the
// children nodes
printNTree(i, flag, depth + 1,
it == (x.root.size()) - 1);
}
flag[depth] = true;
}
// Function to form the Tree and
// print it graphically
static void formAndPrintTree(){
int nv = 10;
tnode r = new tnode(0);
tnode n1 = new tnode(1);
tnode n2 = new tnode(2);
tnode n3 = new tnode(3);
tnode n4 = new tnode(4);
tnode n5 = new tnode(5);
tnode n6 = new tnode(6);
tnode n7 = new tnode(7);
tnode n8 = new tnode(8);
tnode n9 = new tnode(9);
// Array to keep track
// of exploring depths
boolean[] flag = new boolean[nv];
Arrays.fill(flag, true);
// Tree Formation
r.root.add(n1);
n1.root.add(n4);
n1.root.add(n5);
r.root.add(n2);
r.root.add(n3);
n3.root.add(n6);
n3.root.add(n7);
n7.root.add(n9);
n3.root.add(n8);
printNTree(r, flag, 0, false);
}
// Driver Code
public static void main(String[] args)
{
// Function Call
formAndPrintTree();
}
}
// This code is contributed by gauravrajput1 C#
// C# implementation to print
// N-ary Tree graphically
using System;
using System.Collections.Generic;
class GFG
{
// Structure of the node
public class tnode
{
public
int n;
public
List root = new List();
public
tnode(int data)
{
this.n = data;
}
};
// Function to print the
// N-ary tree graphically
static void printNTree(tnode x,
bool[] flag,
int depth, bool isLast )
{
// Condition when node is None
if (x == null)
return;
// Loop to print the depths of the
// current node
for (int i = 1; i < depth; ++i)
{
// Condition when the depth
// is exploring
if (flag[i] == true)
{
Console.Write("| "
+ " "
+ " "
+ " ");
}
// Otherwise print
// the blank spaces
else
{
Console.Write(" "
+ " "
+ " "
+ " ");
}
}
// Condition when the current
// node is the root node
if (depth == 0)
Console.WriteLine(x.n);
// Condition when the node is
// the last node of
// the exploring depth
else if (isLast)
{
Console.Write("+--- " + x.n + '\n');
// No more childrens turn it
// to the non-exploring depth
flag[depth] = false;
}
else
{
Console.Write("+--- " + x.n + '\n');
}
int it = 0;
foreach (tnode i in x.root)
{
++it;
// Recursive call for the
// children nodes
printNTree(i, flag, depth + 1,
it == (x.root.Count) - 1);
}
flag[depth] = true;
}
// Function to form the Tree and
// print it graphically
static void formAndPrintTree()
{
int nv = 10;
tnode r = new tnode(0);
tnode n1 = new tnode(1);
tnode n2 = new tnode(2);
tnode n3 = new tnode(3);
tnode n4 = new tnode(4);
tnode n5 = new tnode(5);
tnode n6 = new tnode(6);
tnode n7 = new tnode(7);
tnode n8 = new tnode(8);
tnode n9 = new tnode(9);
// Array to keep track
// of exploring depths
bool[] flag = new bool[nv];
for(int i = 0; i < nv; i++)
flag[i] = true;
// Tree Formation
r.root.Add(n1);
n1.root.Add(n4);
n1.root.Add(n5);
r.root.Add(n2);
r.root.Add(n3);
n3.root.Add(n6);
n3.root.Add(n7);
n7.root.Add(n9);
n3.root.Add(n8);
printNTree(r, flag, 0, false);
}
// Driver Code
public static void Main(String[] args)
{
// Function Call
formAndPrintTree();
}
}
// This code is contributed by aashish1995 输出
0
+--- 1
| +--- 4
| +--- 5
+--- 2
+--- 3
+--- 6
+--- 7
| +--- 9
+--- 8性能分析：
- 时间复杂度：在上面给出的方法中，有一个递归调用来探索所有需要 O(V) 时间的顶点。因此，这种方法的时间复杂度为O(V) 。
- 辅助空间复杂度：在上述方法中，有额外的空间用于存储探索深度。因此，上述方法的辅助空间复杂度将为O(V)
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